Q.Integrate the following with respect to $x$:

[!FORMULA] $(x+4)^{5}+\dfrac{5}{(2-5x)^{4}}-\text{cosec}^{2}(3x-1)$

Step 1. Split the integral termwise.

Step 2. Integrate the power term. $a=1,\ n=5$:

Step 3. Integrate the reciprocal-power term. Rewrite as $5(2-5x)^{-4}$, $a=-5,\ n=-4$:

so $\displaystyle\int \frac{5}{(2-5x)^4}\,dx = 5\times\frac{1}{15(2-5x)^3}+c=\frac{1}{3(2-5x)^3}+c$.

Step 4. Integrate the $\text{cosec}^2$ term. $a=3$; $\int \text{cosec}^2x\,dx=-\cot x+c$, so

Since this term carries a minus sign in the original expression,

Step 5. Combine all three pieces.

Step 6. Check by differentiating. $\dfrac{d}{dx}\!\left(\dfrac{(x+4)^6}{6}\right)=(x+4)^5$ ✓. $\dfrac{d}{dx}\!\left(\dfrac{1}{3(2-5x)^3}\right)=\dfrac13\cdot(-3)(2-5x)^{-4}\cdot(-5)=5(2-5x)^{-4}=\dfrac{5}{(2-5x)^4}$ ✓. $\dfrac{d}{dx}\!\left(\dfrac13\cot(3x-1)\right)=\dfrac13\cdot\big(-\text{cosec}^2(3x-1)\big)\cdot3=-\text{cosec}^2(3x-1)$ ✓ (matching the original term's minus sign).

Decomposition (§11.5) + linear-argument rule (§11.4) applied termwise

• Losing track of the leading sign carried by the minus in front of the $\text{cosec}^2$ term
• Arithmetic slip combining the outer coefficient $5$ with the $1/a$ correction factor for the middle term